5/6 = 0.833333 .... (non-terminating)
Solving for the polynomial function of least degree with
integral coefficients whose zeros are -5, 3i
We have:
x = -5
Then x + 5 = 0
Therefore one of the factors of the polynomial function is
(x + 5)
Also, we have:
x = 3i
Which can be rewritten as:
x = Sqrt(-9)
Square both sides of the equation:
x^2 = -9
x^2 + 9 = 0
Therefore one of the factors of the polynomial function is (x^2
+ 9)
The polynomial function has factors: (x + 5)(x^2 + 9)
= x(x^2 + 9) + 5(x^2 + 9)
= x^3 + 9x + 5x^2 = 45
Therefore, x^3 + 5x^2 + 9x – 45 = 0
f(x) = x^3 + 5x^2 + 9x – 45
The polynomial function of least degree with integral coefficients
that has the given zeros, -5, 3i is f(x) = x^3 + 5x^2 + 9x – 45
Answer:
Hi. Here is answer if i am not wrong
Answer:
b or c
Step-by-step explanation:
The volume of a cube with side length equal to x, is

,
thus, the volume of a cube shaped box, whose side length is (5a+4b) is :

,
The volume is already expressed in terms of a and b, but we can expand the expression

, as follows:
![(5a+4b)^{3} =(5a+4b)(5a+4b)^{2}= (5a+4b)[ (5a)^{2}+2(5a)(4b)+ (4b)^{2}]](https://tex.z-dn.net/?f=%285a%2B4b%29%5E%7B3%7D%20%3D%285a%2B4b%29%285a%2B4b%29%5E%7B2%7D%3D%20%285a%2B4b%29%5B%20%285a%29%5E%7B2%7D%2B2%285a%29%284b%29%2B%20%284b%29%5E%7B2%7D%5D)
![=(5a+4b)[ 25a^{2}+40ab+ 16b^{2}]](https://tex.z-dn.net/?f=%3D%285a%2B4b%29%5B%2025a%5E%7B2%7D%2B40ab%2B%2016b%5E%7B2%7D%5D)
![=(5a)[ 25a^{2}+40ab+ 16b^{2}]+(4b)[ 25a^{2}+40ab+ 16b^{2}]](https://tex.z-dn.net/?f=%3D%285a%29%5B%2025a%5E%7B2%7D%2B40ab%2B%2016b%5E%7B2%7D%5D%2B%284b%29%5B%2025a%5E%7B2%7D%2B40ab%2B%2016b%5E%7B2%7D%5D)


Answer:

,
or